Exercise 2.4 of Class 9 Maths Chapter 2 - Polynomials introduces the essential concept of the Remainder Theorem. In this exercise, you will learn to use the Remainder Theorem to find the remainder when a polynomial is divided by a linear polynomial. This concept is helpfull when solving larger algebra problems later on.
The Solutions in this exercise are from the NCERT textbooks and practicing them gives you a good chance of scoring better marks in your exams and gives you a clear understanding of the key concepts of polynomials.
To help your learning, we have provided step by step NCERT Solutions for Exercise 2.4 in this article. You can easily download a free PDF of our Practice Solutions to use to practice anytime to develop confidence and be better prepared for exams!
Exercise 2.4 helps you understand the applications of Remainder Theorem for easy calculation. The NCERT Solutions of Class 9 Maths Chapter 2, provides a step by step solution to the questions of this exercise. Download the free PDF of the solutions from below:
1. Use suitable identities to find the following products :
(i) (x+4)(x+10)
(ii) (x+8)(x−10)
(iii) (3x+4)(3x−5)
(iv) (y²+3/2)(y²−3/2)
(v) (3−2x)(3+2x)
Sol. (i) (x+4)(x+10) [using identity (x+a)(x+b) = x²+(a+b)x+ab]
= x²+(4+10)x+(4)(10)
= x²+14x+40
(ii) (x+8)(x−10) [using identity (x+a)(x+b) = x²+(a+b)x+ab]
= (x+8)(x+(−10))
= x²+{8+(−10)}x+8(−10)
= x²−2x−80
(iii) (3x+4)(3x−5) [using identity (x+a)(x+b) = x²+(a+b)x+ab, where x is 3x]
= (3x+4)(3x+(−5))
= (3x)²+{4+(−5)}(3x)+4(−5)
= 9x²−3x−20
(iv) (y²+3/2)(y²−3/2) [using identity (a+b)(a−b)=a²−b²]
Let a=y² and b=3/2.
= (y²)²−(3/2)²
= y⁴−9/4
(v) (3−2x)(3+2x) [using identity (a−b)(a+b)=a²−b²]
= (3)²−(2x)²
= 9−4x²
2. Evaluate the following products without multiplying directly:
(i) 103×107
(ii) 95×96
(iii) 104×96
Sol. (i) 103×107
= (100+3)×(100+7) [using identity (x+a)(x+b) = x²+(a+b)x+ab]
= (100)²+(3+7)(100)+(3)(7)
= 10000+1000+21
= 11021
Alternate solution: 103×107 = (105−2)×(105+2) = (105)²−(2)² = 11025−4 = 11021.
(ii) 95×96
= (90+5)×(90+6) [using identity (x+a)(x+b) = x²+(a+b)x+ab]
= (90)²+(5+6)90+(5)(6)
= 8100+11(90)+30
= 8100+990+30
= 9120
(iii) 104×96
= (100+4)×(100−4) [using identity (a+b)(a−b)=a²−b²]
= (100)²−(4)²
= 10000−16
= 9984
3. Factorise the following using appropriate identities:
(i) 9x²+6xy+y²
(ii) 4y²−4y+1
(iii) x²−y²/100
Sol. (i) 9x²+6xy+y² [using identity (a+b)²=a²+2ab+b²]
= (3x)²+2(3x)(y)+(y)²
= (3x+y)²
= (3x+y)(3x+y)
(ii) 4y²−4y+1 [using identity (a−b)²=a²−2ab+b²]
= (2y)²−2(2y)(1)+(1)²
= (2y−1)²
= (2y−1)(2y−1)
(iii) x²−y²/100 [using identity (a²−b²)=(a+b)(a−b)]
= x²−(y/10)²
= (x+y/10)(x−y/10)
4. Expand each of the following using suitable identities :
(i) (x+2y+4z)²
(ii) (2x−y+z)²
(iii) (−2x+3y+2z)²
(iv) (3a−7b−c)²
(v) (−2x+5y−3z)²
(vi) (1/4 a−1/2 b+1)²
Sol. (Using identity (a+b+c)²=a²+b²+c²+2ab+2bc+2ca)
(i) (x+2y+4z)²
= (x)²+(2y)²+(4z)²+2(x)(2y)+2(2y)(4z)+2(4z)(x)
= x²+4y²+16z²+4xy+16yz+8zx
(ii) (2x−y+z)²
= (2x)²+(−y)²+(z)²+2(2x)(−y)+2(−y)(z)+2(z)(2x)
= 4x²+y²+z²−4xy−2yz+4zx
(iii) (−2x+3y+2z)²
= (−2x)²+(3y)²+(2z)²+2(−2x)(3y)+2(3y)(2z)+2(2z)(−2x) (Note: corrected 2nd last term as per identity)
= 4x²+9y²+4z²−12xy+12yz−8xz
(iv) (3a−7b−c)²
= (3a)²+(−7b)²+(−c)²+2(3a)(−7b)+2(−7b)(−c)+2(−c)(3a) (Note: corrected 2nd last term as per identity)
= 9a²+49b²+c²−42ab+14bc−6ac
(v) (−2x+5y−3z)²
= (−2x)²+(5y)²+(−3z)²+2(−2x)(5y)+2(5y)(−3z)+2(−3z)(−2x)
= 4x²+25y²+9z²−20xy−30yz+12xz
(vi) (1/4 a−1/2 b+1)²
= (1/4 a)²+(−1/2 b)²+(1)²+2(1/4 a)(−1/2 b)+2(−1/2 b)(1)+2(1)(1/4 a)
= (1/16)a²+(1/4)b²+1−(1/4)ab−b+(1/2)a
5. Factorise :
(i) 4x²+9y²+16z²+12xy−24yz−16xz
(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz
Sol. (i) 4x²+9y²+16z²+12xy−24yz−16xz [using identity (a+b+c)²=a²+b²+c²+2ab+2bc+2ca]
Notice the negative terms involving z. This means z must be negative in the factored form.
= (2x)²+(3y)²+(−4z)²+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)
= (2x+3y−4z)²
= (2x+3y−4z)(2x+3y−4z)
(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz [using identity (a+b+c)²=a²+b²+c²+2ab+2bc+2ca]
Notice the negative terms involving x and xz. This means x must be negative in the factored form.
= (−√2x)²+(y)²+(2√2z)²+2(−√2x)(y)+2(y)(2√2z)+2(2√2z)(−√2x)
= (−√2x+y+2√2z)²
6. Write the following cubes in expanded form :
(i) (2x+1)³
(ii) (2a−3b)³
(iii) (3/2 x+1)³
(iv) (x−2/3 y)³
Sol. (Using identity (a+b)³=a³+b³+3ab(a+b) and (a−b)³=a³−b³−3ab(a−b))
(i) (2x+1)³
= (2x)³+(1)³+3(2x)(1)(2x+1)
= 8x³+1+6x(2x+1)
= 8x³+1+12x²+6x
= 8x³+12x²+6x+1
(ii) (2a−3b)³
= (2a)³−(3b)³−3(2a)(3b)(2a−3b)
= 8a³−27b³−18ab(2a−3b)
= 8a³−27b³−36a²b+54ab²
(iii) (3/2 x+1)³
= (3/2 x)³+(1)³+3(3/2 x)(1)(3/2 x+1)
= (27/8)x³+1+(9/2)x(3/2 x+1)
= (27/8)x³+1+(27/4)x²+(9/2)x
= (27/8)x³+(27/4)x²+(9/2)x+1
(iv) (x−2/3 y)³
= x³−(2/3 y)³−3x(2/3 y)(x−2/3 y)
= x³−(8/27)y³−2xy(x−2/3 y)
= x³−(8/27)y³−2x²y+(4/3)xy²
7. Evaluate the following using suitable identities :
(i) (99)³
(ii) (102)³
(iii) (998)³
Sol. (i) (99)³
= (100−1)³ [using (a−b)³=a³−b³−3ab(a−b)]
= (100)³−(1)³−3(100)(1)(100−1)
= 1000000−1−300(99)
= 1000000−1−29700
= 970299
(ii) (102)³
= (100+2)³ [using (a+b)³=a³+b³+3ab(a+b)]
= (100)³+(2)³+3(100)(2)(100+2)
= 1000000+8+600(102)
= 1000000+8+61200
= 1061208
(iii) (998)³
= (1000−2)³ [using (a−b)³=a³−b³−3ab(a−b)]
= (1000)³−(2)³−3(1000)(2)(1000−2)
= 1000000000−8−6000(998)
= 1000000000−8−5988000
= 994011992
8. Factorise each of the following :
(i) 8a³+b³+12a²b+6ab²
(ii) 8a³−b³−12a²b+6ab²
(iii) 27−125a³−135a+225a²
(iv) 64a³−27b³−144a²b+108ab² (Note: the original text had 108a² which seems like a typo, assuming 108ab² for consistency with (a-b)³)
(v) 27p³−1/216−(9/2)p²+(1/4)p
Sol. (i) 8a³+b³+12a²b+6ab² [using identity (a+b)³=a³+b³+3a²b+3ab² = a³+b³+3ab(a+b)]
= (2a)³+(b)³+3(2a)²(b)+3(2a)(b)²
= (2a)³+(b)³+3(2a)(b)(2a+b)
= (2a+b)³
= (2a+b)(2a+b)(2a+b)
(ii) 8a³−b³−12a²b+6ab² [using identity (a−b)³=a³−b³−3a²b+3ab²]
= (2a)³+(−b)³+3(2a)²(−b)+3(2a)(−b)²
= (2a−b)³
(iii) 27−125a³−135a+225a² [Rearrange to a³−b³−3a²b+3ab² form: (3)³−(5a)³+225a²−135a ]
= (3)³−(5a)³+3(3)(5a)(−5a+3) (incorrect grouping in solution)
Correct interpretation: 27−125a³−135a+225a² = 3³−(5a)³−3(3)²(5a)+3(3)(5a)²
= 3³−(5a)³−3(9)(5a)+3(3)(25a²)
= 3³−(5a)³−135a+225a²
This matches the form of (a−b)³ = a³−b³−3a²b+3ab².
So, it is (3−5a)³.
(iv) 64a³−27b³−144a²b+108ab² [using identity (a−b)³=a³−b³−3a²b+3ab²]
= (4a)³−(3b)³−3(4a)²(3b)+3(4a)(3b)²
= (4a)³−(3b)³−3(16a²)(3b)+3(4a)(9b²)
= (4a)³−(3b)³−144a²b+108ab²
So, it is (4a−3b)³.
(v) 27p³−1/216−(9/2)p²+(1/4)p [using identity (a−b)³=a³−b³−3a²b+3ab²]
= (3p)³−(1/6)³−3(3p)²(1/6)+3(3p)(1/6)²
= (3p)³−(1/6)³−3(9p²)(1/6)+3(3p)(1/36)
= (3p)³−(1/6)³−(27/6)p²+(9/36)p
= (3p)³−(1/6)³−(9/2)p²+(1/4)p
So, it is (3p−1/6)³.
= (3p−1/6)(3p−1/6)(3p−1/6)
Verify:
(i) x³+y³=(x+y)(x²−xy+y²)
(ii) x³−y³=(x−y)(x²+xy+y²)
Sol. (i) We know (x+y)³=x³+y³+3xy(x+y)
Rearranging for x³+y³:
x³+y³=(x+y)³−3xy(x+y)
Factor out (x+y):
x³+y³=(x+y)[(x+y)²−3xy]
Expand (x+y)²:
x³+y³=(x+y)(x²+2xy+y²−3xy)
Simplify:
x³+y³=(x+y)(x²−xy+y²) (Verified)
(ii) We know (x−y)³=x³−y³−3xy(x−y)
Rearranging for x³−y³:
x³−y³=(x−y)³+3xy(x−y)
Factor out (x−y):
x³−y³=(x−y)[(x−y)²+3xy]
Expand (x−y)²:
x³−y³=(x−y)(x²+y²−2xy+3xy)
Simplify:
x³−y³=(x−y)(x²+y²+xy) (Verified)
9. Factorise each of the following :
(i) 27y³+125z³
(ii) 64m³−343n³
Sol. (i) 27y³+125z³ [using identity a³+b³=(a+b)(a²−ab+b²)]
= (3y)³+(5z)³
= (3y+5z){(3y)²−(3y)(5z)+(5z)²}
= (3y+5z)(9y²−15yz+25z²)
(ii) 64m³−343n³ [using identity a³−b³=(a−b)(a²+ab+b²)]
= (4m)³−(7n)³
= (4m−7n){(4m)²+(4m)(7n)+(7n)²}
= (4m−7n)(16m²+28mn+49n²)
Factorise: 27x³+y³+z³−9xyz
Sol. 27x³+y³+z³−9xyz [using identity a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca)]
Here, a=3x, b=y, c=z.
= (3x)³+(y)³+(z)³−3(3x)(y)(z)
= (3x+y+z)((3x)²+(y)²+(z)²−(3x)(y)−(y)(z)−(z)(3x))
= (3x+y+z)(9x²+y²+z²−3xy−yz−3zx)
Verify that x³+y³+z³−3xyz = (1/2)(x+y+z) [(x−y)²+(y−z)²+(z−x)²]
Sol. Start from the R.H.S.:
(1/2)(x+y+z)[(x−y)²+(y−z)²+(z−x)²]
Expand the squares:
= (1/2)(x+y+z)[(x²−2xy+y²)+(y²−2yz+z²)+(z²−2zx+x²)]
Combine like terms inside the square brackets:
= (1/2)(x+y+z)[2x²+2y²+2z²−2xy−2yz−2zx]
Factor out 2 from the square brackets:
= (1/2)(x+y+z)2(x²+y²+z²−xy−yz−zx)
Cancel the 1/2 and 2:
= (x+y+z)(x²+y²+z²−xy−yz−zx)
This is the standard identity for x³+y³+z³−3xyz.
= x³+y³+z³−3xyz (Verified)
If x+y+z=0, show that x³+y³+z³=3xyz.
Sol. We know the identity:
x³+y³+z³−3xyz=(x+y+z)(x²+y²+z²−xy−yz−zx)
Given: x+y+z=0. Substitute this into the identity:
x³+y³+z³−3xyz = (0)(x²+y²+z²−xy−yz−zx)
x³+y³+z³−3xyz = 0
Therefore, x³+y³+z³=3xyz.
10. Without actually calculating the cubes, find the value of each of the following :
(i) (−12)³+(7)³+(5)³
(ii) (28)³+(−15)³+(−13)³
Sol. (Using the identity: if a+b+c=0, then a³+b³+c³=3abc)
(i) Let a=−12, b=7, c=5.
Check if a+b+c=0: −12+7+5 = −12+12 = 0.
Since a+b+c=0, then a³+b³+c³=3abc.
(−12)³+(7)³+(5)³ = 3(−12)(7)(5)
= 3(−420) = −1260.
(ii) Let a=28, b=−15, c=−13.
Check if a+b+c=0: 28+(−15)+(−13) = 28−15−13 = 28−28 = 0.
Since a+b+c=0, then a³+b³+c³=3abc.
(28)³+(−15)³+(−13)³ = 3(28)(−15)(−13)
= 3(28)(195)
= 84 × 195 = 16380.
11. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :
(i) Area = 25a²−35a+12
(ii) Area = 35y²+13y−12
Sol. (Area of rectangle = Length × Breadth. Factorise the quadratic expression)
(i) Area = 25a²−35a+12
To factorise, find two numbers whose product is 25×12=300 and sum is -35. These are -20 and -15.
= 25a²−20a−15a+12
= 5a(5a−4)−3(5a−4)
= (5a−3)(5a−4)
Here, Length = 5a−3, Breadth = 5a−4 (or vice versa).
(ii) Area = 35y²+13y−12
To factorise, find two numbers whose product is 35×(−12)=−420 and sum is 13. These are 28 and -15.
= 35y²+28y−15y−12
= 7y(5y+4)−3(5y+4)
= (5y+4)(7y−3)
Here, Length = 5y+4, Breadth = 7y−3 (or vice versa).
12. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: 3x²−12x
(ii) Volume: 12ky²+8ky−20k
Sol. (Volume of cuboid = Length × Breadth × Height. Factorise the expression into three factors)
(i) Volume = 3x²−12x
Factor out the common term 3x:
= 3x(x−4)
So, the dimensions are 3 units, x units and (x−4) units.
(ii) Volume = 12ky²+8ky−20k
Factor out the common term 4k:
= 4k(3y²+2y−5)
Now factor the quadratic expression (3y²+2y−5). Find two numbers whose product is 3×(−5)=−15 and sum is 2. These are 5 and -3.
= 4k(3y²+5y−3y−5)
= 4k{y(3y+5)−1(3y+5)}
= 4k(3y+5)(y−1)
So, the dimensions of the cuboid are 4k, (3y+5), and (y−1).
(Session 2025 - 26)